hdu4300 Clairewd’s message(字符串匹配 hash)

Clairewd’s message

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 8094    Accepted Submission(s): 2982

Problem Description

Clairewd is a member of FBI. After several years concealing in BUPT, she intercepted some important messages and she was preparing for sending it to ykwd. They had agreed that each letter of these messages would be transfered to another one according to a conversion table.
Unfortunately, GFW(someone's name, not what you just think about) has detected their action. He also got their conversion table by some unknown methods before. Clairewd was so clever and vigilant that when she realized that somebody was monitoring their action, she just stopped transmitting messages.
But GFW knows that Clairewd would always firstly send the ciphertext and then plaintext(Note that they won't overlap each other). But he doesn't know how to separate the text because he has no idea about the whole message. However, he thinks that recovering the shortest possible text is not a hard task for you.
Now GFW will give you the intercepted text and the conversion table. You should help him work out this problem.

 

 

Input

The first line contains only one integer T, which is the number of test cases.
Each test case contains two lines. The first line of each test case is the conversion table S. S[i] is the ith latin letter's cryptographic letter. The second line is the intercepted text which has n letters that you should recover. It is possible that the text is complete.

Hint

Range of test data:
T<= 100 ;
n<= 100000;

 

 

Output

For each test case, output one line contains the shorest possible complete text.

 

 

Sample Input

2 abcdefghijklmnopqrstuvwxyz abcdab qwertyuiopasdfghjklzxcvbnm qwertabcde

 

 

Sample Output

abcdabcd qwertabcde

 

 

Author

BUPT

 

 

Source

2012 Multi-University Training Contest 1

 

 

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关于字符串匹配(Hash算法), 可以看一下https://www.cnblogs.com/zyf0163/p/4806951.html

题目大意:

是有一份文件，前面是密文，后面是原文，但那个人接到这个文件后不知道中间从哪里开始是原文，所以你要帮忙还原一下，如果后面原文比密文少，你就将它补全, 第一行是密文转换格式，例如第二个样例表示将q翻译成a,w翻译成b。

思路：

我们只要先把密文都翻译成明文，然后去比较原来的字符串的后缀和翻译之后的字符串前缀的最长匹配长度就行（注：最长匹配的长度不能超过原长的一半）

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
using namespace std;

typedef unsigned long long ull;
const int N = 1000005;
const int seed = 163;
char a[N], b[N];
int c[N];

int T;
ull p[N];

vector<ull>v1, v2;

void init(){
    p[0] = 1;
    for(int i=1; i<=100000; i++)
        p[i] = p[i-1] * seed;
}

ull rkhash(char *s, int id){
    ull Hash = 0;
    for(int i=1; s[i]; i++){
        if(id == 1){
            Hash = (Hash*seed + s[i]-'a');
            v1.push_back(Hash);
        }else if(id == 2){
            Hash = (Hash*seed + c[s[i]-'a']);
            v2.push_back(Hash);
        }
    }
    return Hash;
}

ull get(int l, int r, vector<ull> &g){
    return g[r] - g[l-1] * p[r-l+1];
}

void work(){
    v1.clear(); v2.clear();
    v1.push_back(0); v2.push_back(0);
    for(int i=0; i<26; i++) c[a[i]-'a'] = i;
    rkhash(b, 1); rkhash(b, 2);
    int n = strlen(b+1);
    int ans = n;
    for(int i=n; i<n*2; i++){
        if(i & 1) continue;
        int tmp = i / 2;
        int len = n-tmp;
        ull s1 = get(1, len, v2);
        ull s2 = get(n-len+1, n, v1);
        if(s1 == s2){
            ans = tmp;
            break;
        }
    }
    for(int i=1; i<=ans; i++) printf("%c", b[i]);
    for(int i=1; i<=ans; i++) printf("%c", c[b[i]-'a']+'a');
    puts("");
}

int main(){
    scanf("%d", &T);
    init();
    
    while(T--){
        scanf("%s%s", a, b+1);
        work();
    }
    

    return 0;
}

 当然还有KMP的做法， 不过我不会， 等将来填坑.......